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From a collection of 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are chosen and arranged in a row such that the dictionary always occupies the middle position. How many such arrangements exist?
Choose 4 novels from 6: $\binom{6}{4} = 15$ ways.
Choose 1 dictionary from 3: $\binom{3}{1} = 3$ ways.
Arrange 5 books in a row with dictionary fixed in the middle: the 4 novels occupy the 4 remaining positions, giving $4! = 24$ arrangements.
Total $= 15 \times 3 \times 24 = \mathbf{1080}$
Since $1080 \geq 1000$, the answer is at least 1000.
If $\cos^{-1}x - \cos^{-1}\!\dfrac{y}{2}=\alpha$ where $-1\leq x\leq1,\ -2\leq y\leq2,\ x\leq\dfrac{y}{2}$, then $4x^2-4xy\cos\alpha+y^2$ equals:
Let $\cos^{-1}x=A$ and $\cos^{-1}(y/2)=B$, so $A-B=\alpha$.
Then $\cos A=x,\ \cos B=y/2,\ \sin A=\sqrt{1-x^2},\ \sin B=\sqrt{1-y^2/4}$.
Using $\cos\alpha=\cos(A-B)=\cos A\cos B+\sin A\sin B=\dfrac{xy}{2}+\sqrt{1-x^2}\sqrt{1-\frac{y^2}{4}}$
$4x^2-4xy\cos\alpha+y^2 = (2x-y\cos\alpha)^2+y^2(1-\cos^2\alpha)$... expanding directly: $= 4x^2+y^2-4xy\cos\alpha$.
After substitution and simplification using the expression for $\cos\alpha$: $= 4(1-\cos^2\alpha) = \mathbf{4\sin^2\alpha}$.
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